Physics for You - May 2016

Volume 24 Managing Editor Mahabir Singh

May 2016

Corporate Office: Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (HR). Tel : 0124-4951200 e-mail : [email protected] website : www.mtg.in Regd. Office: 406, Taj Apartment, Near Safdarjung Hospital, New Delhi - 110029.

CONTENTS

Editor Anil Ahlawat (BE, MBA)

No. 5

Physics Musing Problem Set 34

8

JEE Advanced Practice Paper

10

JEE Main Solved Paper 2016

22

PMT Practice Paper

31

BITSAT Practice Paper

39

Brain Map

46

Exam Prep 2016

51

AIIMS Practice Paper

56

Olympiad Problems

65

JIPMER Practice Paper

69

Core Concept

74

Live Physics

81

Physics Musing Solution Set 33

82

You Ask We Answer

84

Crossword

85

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Send D.D/M.O in favour of MTG Learning Media (P) Ltd. Payments should be made directly to : MTG Learning Media (P) Ltd, Plot No. 99, Sector 44, Gurgaon - 122003 (Haryana) We have not appointed any subscription agent. Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and printed by Personal Graphics and Advertisers (P) Ltd., Okhla Industrial Area, Phase-II, New Delhi. Readers are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine. Focus/Infocus features are marketing incentives. MTG does not vouch or subscribe to the claims and representations made by advertisers. All disputes are subject to Delhi jurisdiction only. Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd. All rights reserved. Reproduction in any form is prohibited.

Physics For you | may ‘16

7

P

PHYSICS

MUSING

hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material. In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of these problems will be published in next issue of Physics For You. The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those who send atleast five correct solutions will be published in the next issue. We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams.

1. A particle is projected from ground in vertical direction at t = 0. At t = 0.8 s, it reaches h = 14 m. It will again come to same height at t = (g = 10 m s–2) 14 7 s (c) 3 s (a) 2 s (b) (d) s 5 2 2. In the figure shown, the blocks A and C are pulled down with constant velocities u . Acceleration of block B is b

A

(a)

u2 tan2 θ sec θ b

u

b

 

B

(b)

C

u

u2 tan3 θ b

u2 2 (d) zero sec θ tan θ b 3. Two particles A and B each of mass m are attached by a light inextensible string of length 2l. The whole system lies on a smooth horizontal table with B initially at a distance l from A. The particle at end B is projected across the table with speed u perpendicular to AB. Velocity of ball A just after the jerk is u 3 (a) (b) u 3 4 u u 3 (c) (d) 2 2 4. A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere of radius r as shown in the figure. The particle looses its contact with hemisphere at point B. C is centre of the hemisphere. The equation relating a and b is (c)

A B



r C



(a) 3 sin a = 2 cos b (b) 2 sin a = 3 cos b (c) 3 sin b = 2 cos a (d) 2 sin b = 3 cos a 5. The intensity of radiation emitted by the Sun has its maximum value at a wavelength of 510 nm and that emitted by the North star has the maximum value at 350 nm. If these stars behave like black bodies, then the ratio of the surface temperatures of the Sun and the North star is (a) 1.46 (b) 0.69 (c) 1.21 (d) 0.83 6. N(< 100) molecules of a gas have velocities 1, 2, 3........ N km s–1 respectively. Then (a) rms speed and average speed of molecules is same (b) ratio of rms speed to average speed of molecules is (2N + 1)(N + 1) / 6N (c) ratio of rms speed to average speed of molecules is (2N + 1)(N + 1) / 6 (d) ratio of rms speed to average speed of molecules (2N + 1) 6(N + 1) 7. A thermodynamic process of one mole ideal gas is shown in the V 4T0 figure. The efficiency 2V C 0 of cyclic process ABCA will be A B V0 T0 (a) 25% 2T0 (b) 12.5% 2P0 P P0 (c) 50% (d) 7.7% Contd. on page no. 80 is 2

By Akhil Tewari, author Foundation of Physics for JEE main & advanced, Senior Professor Physics, RaO IIT aCaDEmy, mumbai.

8

Physics For you | may ‘16

JEE Advanced exam on 22nd May 2016

PRACTICE PRACTICE PAPER PAPER 2016 2016 PaPer-1

Section 1 (MaxiMuM MarkS : 32)

• •

This section contains EIGHT questions The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive

1. A current I flows in a rectangularly shaped wire whose center lies at (x0, 0, 0) and whose vertices are located at the points A(x0 + d, –a, –b), B(x0 – d, a, –b), C(x0 – d, a, b), and D(x0 + d, –a, b) respectively. Assume that a, b, d R (Outside point) where, q =

= Magnetic force

Cyclotron Accelerates charged particles so they acquire high energies,

Broadside-on,

Biot Savart law Ampere's circuital law

Moving Coil Galvanometer Current sensitivity,

Net magnetic field, Here, BH = B cos d, BV = B sin d d is the dip angle at the place.

Applications of Ampere's circuital law

Applications of BiotSavart law

Magnetic field due to a straight wire of infinite length,

Field on the axis of a circular loop

Curie law, Magnetic field inside a long solenoid, B = µ0nI

Curie-Weiss law,

Ohm's law V = IR, Resistivity Heat energy, H = I 2Rt P = I 2R =

For transmission cable, power loss

Drift speed, I = neAvd ,

Temperature dependence RT = R0[1 + a(T – T0)] Temperature coefficient of resistance,

Kirchhoff’s rules Junction Rule : Current entering = Current leaving, Loop Rule : Total potential around any closed loop must be zero, Potentiometer Fall of potential , Vµ l , V = Kl Here,

Voltage sensitivity,

Galvanometer into voltmeter

Field at the centre of a loop

Based on principle of Electromagnetic Induction

Power,

Magnetic field due to a bar magnet End-on,

fc

Earth’s magnetism and magnetic properties of substances

sphere

Faraday law of EMI

Electric generator and electric motor

Flux linked, f = NBA coswt Instantaneous induced emf, e = e0sinwt

e = e0 sinwt Lenz's law T h e d i re c t i o n o f t h e induced current is such that it opposes the change that has induced it. Eddy current The current induced in conductors when the magnetic flux linked with the conductor changes.

Motional emf On a straight conducting wire, e = Bvl On a rotating conducting wire about one end

Meter bridge Resistance of unknown resistor,

Self inductance of a long solenoid

Mutual inductance of two long coaxial solenoids

Transformer Efficiency of transformer,

When efficiency of transformer is 100%,

Emf induced in the coil/conductor,

= k (transformer ratio)

Coefficient of self induction,

Wheatstone bridge In balanced condition,

Galvanometer into ammeter

Energy stored in an inductor

Mutual inductance,

Series LCR circuit Impedance of series LCR circuit

= potential gradient

Electric resonance Comparision of emfs of two cells,

Internal resistance of a cell

At resonance, XL = XC \ Resonance frequency

Power factor

Q-factor of LCR circuit

So, the rate of flow of water in venturimeter is given by 2 gh V = A1A2 2 (A1 − A22) = 6.25p × 16p =

2 × 980 × 4 (16p)2 − (6.25p)2

100p 2 × 28 10 (16p − 6.25p)(16p + 6.25p)

= 1889 cc s −1

14. (b) : As we know according to equation of continuity, when cross-section of duct decreases, the velocity of flow of liquid increases and in accordance with Bernoulli’s theorem, in a horizontal pipe, the place where speed of liquid is maximum, the value of pressure is minimum. Hence the second graph correctly represents the variation of pressure. 15. (a) : Here, A = 25 cm, T = 3 s Let the particle be at the location – 12.5 cm at time t1 and + 12.5 cm at time t2.  2pt  Using the relation, x = A cos  + f  T  2pt First condition, −12.5 = 25 cos  1 + f  ...(i)  3   2pt 2  Second condition, 12.5 = 25 cos  + f ...(ii)  3  2pt 1 2p −12.5 From eq. (i), cos  1 + f  = = − = cos 25 2 3  3  2pt1 2p \ +f= 3 3 or 2pt1 + 3f = 2p ...(iii) p  2pt 2  12.5 1 = = cos + f = From eq. (ii), cos  25 2 3  3  2pt 2 p \ +f= or 2pt2 + 3f = p ...(iv) 3 3 Subtracting eq. (iv) from eq. (iii), we get 2p(t1 – t2) = p or (t1 – t2) = p/2p = 1/2 = 0.5 s 16. (c) : If q = charge on the drop, then q = ne = 12 × 1.6 × 10–19 C = 19.2 × 10–19 C If Fe be the electrostatic force on the oil drop due to electric field. Then Fe = qE = 19.2 × 10–19 × 2.55 × 104 N ...(i) Let Fg = Force on the drop due to gravity, then 4 Fg = mg = pr 3rg ...(ii) 3 48

Physics For you | may ‘16

Here r = density of oil = 1.26 g cm–3 = 1.26 × 103 kg m–3 Putting these values in eq. (ii), we get 4 ...(iii) Fg = pr 3 × 1.26 × 103 × 9.81 3 As the drop remains stationary, Fe = Fg 4 or 19.2 × 10–19 × 2.55 ×104 = pr 3 × 1.26 × 103 × 9.81 3 After solving we get, r = 9.82 × 10–4 mm 17. (d) : Here q = charge at origin O = 8 mC = 8 × 10–3 C q0 = charge to be carried from P to Q via R = – 2 × 10–9 C \ r1 = 3 cm = 3 × 10–2 m r2 = 4 cm = 4 × 10–2 m As electrostatic forces are conservative forces, the work done in moving q0 is independent of the path followed. Thus there is no relevance of the point R. Let WPQ be the work done in moving q0 from P to Q, then using the relation, 1 1 1 WPQ = qq − , we get 4pε 0 0  r2 r1  WPQ = 9 × 109 × (– 2 × 10–9) 1   1 × 8 × 10 −3  −  = 1.2 J − 2  4 × 10 3 × 10 −2  18. (d) : In parallel, current distributes in inverse ratio of resistance. 19. (b) : When final image is formed at infinity, then magnifying power of compound microscope v D 1 1 1 M ∞ = − 0 × . From, = − f 0 v 0 u0 u0 fe 1 1 1 = − ⇒ v0 = 30 cm + 1.2 v 0 (−1.25) \ | M∞ | =

30 25 × = 200 1.25 3

20. (a) : Dm = 2 (2.015) – (3.017 + 1.009) = 0.004 amu \ Energy released = (0.004 × 931.5) MeV = 3.726 MeV 3.726 Energy released per deuteron = = 1.863 MeV 2 Number of deuterons in 1 kg = \

6.02 × 1026 = 3.01 × 1026 2

Energy released per kg of deuterium fusion = (3.01 × 1026 × 1.863) = 5.6 × 1026 MeV ≈ 9.0 × 1013 J

21. (c) : Velocity of sound in a gas, v = For the same temperature

γRT M

γ v∝ M

7 ×4 42 5 = \ 5 5 ×2 3 22. (b) : Among the given physical quantities pressure, power, energy, gravitational potential, electrical charge, temperature are scalar quantities whereas only impulse and acceleration are vector quantities. γH M He 2 2 = = γ He MH vHe 2

vH

23. (c) 24. (b) : Let original frequency of sitar string A be uA and original frequency of sitar string B be uB. Number of beats per second = 6 \ uB = uA ± 6 = 324 ± 6 = 330 or 318 Hz When tension in A is reduced, its frequency reduces ( u ∝ T ) Number of beats per second reduces to 3. Therefore, frequency of B = 324 – 6 = 318 Hz. 25. (b) 26. (d) : Volume of the cylinder, V = pr2L Volumetric strain =

DV D(pr 2 L) = V pr 2 L

2 DV pr DL + 2 pr L Dr DL 2Dr = + = L r V pr 2 L

... (i)

(Dr / r ) Dr sDL or =− (DL / L) r L Dr On substituting this value of in eq. (i), we get r DV DL ... (ii) = (1 − 2s) V L (F / pr 2 ) DL F Young’s modulus, Y = or = (DL / L) L pr 2Y DL On substituting this value of in eq. (ii), we get L DV F DV F (1 − 2s) , = = (1 − 2s) 2 2 V pr Y pr L pr 2Y

Poisson’s ratio, s = −

DV =

FL (1 − 2s) Y

r 2r and r2 = 3 3 w will be same for both the stars. 2m 1 1 K1 = I1w2 and K 2 = I2 w2 2 2 2  2r  m   K1 I1 mr12 3 \ = = = =2 2 2 K 2 I2 2mr2 r  2m   3 L1 = I1w and L2 = I2w

27. (d) : r1 =

r2

r1 CM r

L1 I1 = =2 L2 I2 28. (b) : At temperature T, energy is maximum around the wavelength l0. According to Wien’s law b l0T = b or T = ... (i) l0 \

Power radiated by a black body at temperature T is 4

 b  P = sT A = s   A (Using (i))  l0  b4 ...(ii) =s A l04 Now at new temperature T′, energy is maximum l around wavelength 0 . 2 l0 2b ...(iii) T ′ = b or T ′ = \ 2 l0 Now power radiated by the same black body at temperature T′ is 4

4

 2b  P ′ = sT ′ A = s   A  l0  b4 P ′ = 16 s A = 16 P l04 4

(Using (iii)) (Using (ii))

29. (c) : For SHM, Maximum velocity, vmax = Aw k For 1st case, vmax = A1w1 = A1 1 1 M

...(i)

k For IInd case, vmax = A2 w2 = A2 2 2 M v = v max max According to question, 1

\ A1

k1 k = A2 2 M M

or

...(ii)

2

A1 k = 2 A2 k1

Physics For you | may ‘16

49

30. (b) : In figures (1), (3) and (4) net electric field is zero, because electric field at a point due to positive charge acts away from the charge and due to negative charge it acts towards the charge. For figure, (2) net electric field is not zero. E

E 120° E

Here, net electric field in figure, (2) is (2E)2 + (2E)2 + (2E)2 ⋅ 2 cos 120° = 2E 31. (d) : 30°

60°

If q is the angle which the velocity v of electron while leaving the electric field makes with the initial direction of motion (i.e., along X-direction), then vy y tan q = = vx d vy eE Ld Ee L 1 or y = d = × × d= vx m vx vx mv 2x 39. (d) : For pure inductor, ε 100 25 XL = u = = Ω Iu 8 2 25 25 25 1 wL = ; L = = = H 2 2w 2 × 2 × p × 50 8 p V 100 R= = = 10 Ω I 10 For the combination, the supply is 150 V, 40 Hz 1 \ XL = wL = 2p × 40 × = 10 Ω 8p

90° 60° 60°

Using snell’s law at the interface, sin 60° 3 /2 m= = = 3 sin 30° 1 / 2 32. (a) : Path difference between two waves yd  1.25lD   d  = D  d   D  5 = 1.25l = l 4 2p Corresponding phase difference, f = ⋅ Dx l 2p 5 5 f 5 = × l = p , = p = 225° l 4 2 2 4 f 1 I = I max cos 2 = I max 2 2 Dx = d sin q  d tan q =

33. (a) : 92 U

238

a

 →90 Th

234

b

→  91 Pa

234

−1 b

0

 →92 U

234

So, A = 234, B = 90, C = 234, D = 91 and E = b 34. (b) 35. (b) 36. (c) : In reflected light, condition for maximum intensity is l 2mt cos r = (2n + 1) 2 4mt cos r 4 × 1.33 × 5 × 10 −7 × 1 \ l= = 2n + 1 (2n + 1) Putting n = 0, 1, 2, 3, ... we get l0 = 26600 Å, l1 = 8866 Å l2 = 5320 Å, l4 = 3800 Å Among these wavelengths, l = 5320 Å is in the visible region. 50

Physics For you | may ‘16

37. (c) 38. (b) : When electron is subjected to a perpendicular electric field, it will describe a parabolic path in the electric field. While leaving the electric field, it flies off tangentially to the parabolic path and meets the screen at point P, where MP = y. Let vx, vy be the rectangular components of velocity when the electron just going to leave the electric field. If t is the time taken by electron to cross the electric field, then vx × t = L or t = L/vx Ee Ee L vy = 0 + t = × m m vx

Z = X L2 + R2 = 102 + 102 = 10 2 Ω E 150 15 Iu = u = = A Z 10 2 2 40. (b) : If m is strength of each pole of wire of length l, A then M = m × l When the wire is bent r x such that ∠AOB = 60° 30° l O 30° 2 pr pr x l= = r 6 3 pr B \ M =m× 3 In the figure linear distance between A and B = 2x = 2 r sin30° = r 3M \ M ′ = m(2 x ) = mr = p nn

chapterwise McQs for practice

Useful for All National and State Level Medical/Engg. Entrance Exams alternating current

1. An alternating emf is applied across a parallel combination of a resistance R, capacitance C and an inductance L. If IR, IL and IC are the currents through R, L and C respectively, then the diagram which correctly represents the phase relationship among IR, IL, IC and source emf e, is given by IR

IL

(a)

IR

 (b)

IL

IC

IC

IC

IR

(c)

IR IL

 (d)

IC





IL

2. In the circuit shown, the AC source has voltage V = 20 cos(wt) volt with w = 2000 rad s–1, the amplitude of the current will be nearest to

(a) 2 A (c) 2/ 5 A

(b) 3.3 A (d) 5 A

3. The voltage across a pure inductor is represented in figure. Which one of the following curves in the figure will represent the current ?

(a)

(b)

(c)

(d)

4. A direct current of 2 A and an alternating current having a maximum value of 2 A flow through two identical resistances. The ratio of heat produced in the two resistances will be (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 4 : 1 5. A 750 Hz, 20 V source is connected to a resistance of 100 W, an inductance of 0.1803 H and a capacitance of 10 µF all in series. Calculate the time in which the resistance (thermal capacity 2 J°C–1) will get heated by 10°C. (a) 348 s (b) 353 s (c) 365 s (d) 370 s 6. A power transmission line feeds input power at 2300 V to a step-down transformer, with its primary windings having 4000 turns. What should be the number of turns in the secondary winding in order to get output power at 230 V ? (a) 300 (b) 250 (c) 400 (d) 450 7. An inductance and a resistance are connected in series with an AC potential. In this circuit (a) the current and the potential difference across the resistance lead the potential difference across the inductance by phase angle p/2 (b) the current and the potential difference across the resistance lag behind potential difference across the inductance by an angle p/2 physics for you | MAY ‘16

51

(c) the current and the potential difference across the resistance lag behind the potential difference across the inductance by an angle p (d) the potential difference across the resistance lags behind the potential difference across the inductance by an angle p/2 but the current in the resistance leads the potential difference across inductance by p/2. 8. An AC source is connected to a capacitor. The current in the circuit is I. Now a dielectric slab is inserted into the capacitor, then the new current is (a) equal to I (b) more than I (c) less than I (d) may be more than or less than I 9. In the given AC circuit

(a) (b) (c) (d)

current I2 and V are in same phase current I2 leads I1 by 90° current I leads I2 by q < 90° current I leads I1 by q < 90°

10. If the rms current in a 50 Hz AC circuit is 5 A, the value of the current 1/300 seconds after its value becomes zero is (a) 5 2 A (b) 5 3 / 2 A (c) 5/6 A (d) 5 / 2 A 11. A transformer is used to light 140 W, 24 V lamp from 240 V AC mains. The current in the mains is 0.7 A. The efficiency of transformer is nearest to (a) 90% (b) 80% (c) 70% (d) 60% 12. An L-C circuit has capacitance C1 = C and inductance C and L2 = 2L L1 = L. A second circuit has C2 = 2 and a third circuit has C3 = 2C and L3 = L/2. All the three capacitors are charged to the same potential V, and then made to oscillate. Then (a) maximum current is greatest in second circuit (b) angular frequency of oscillation is different for all the three circuits. (c) maximum current is greatest in first circuit (d) angular frequency of oscillation is same for all the three circuits. 52

physics for you | MAY ‘16

13. An alternating voltage is given by e = e1 sinwt + e2 coswt Then the root mean square value of voltage is given by (a) e12 + e22 (b) e1 + e 2 e1e2 e12 + e22 (d) 2 2 14. An AC voltage source of variable angular frequency w and fixed amplitude V0 is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When w is increased (a) the bulb glows dimmer (b) the bulb glows brighter (c) total impedance of the circuit is unchanged (d) total impedance of the circuit increases. (c)

15. An electric bulb has a rated power of 50 W at 100 V. If it is used on an AC source of 200 V, 50 Hz, a choke has to be used in series with it. This choke should have an inductance of (a) 1 mH (b) 0.1 mH (c) 0.1 H (d) 1.1 H 16. When a DC voltage of 200 V is applied to a coil of self inductance (2 3 / p) H, a current of 1 A flows through it. But by replacing DC source with AC source of 200 V, the current in the coil is reduced to 0.5 A. Then the frequency of AC supply is (a) 100 Hz (b) 75 Hz (c) 50 Hz (d) 30 Hz 17. For the circuit shown in figure, the current through the inductor is 0.9 A while the current through the condenser is 0.4 A. Hence the current drawn from the generator is (a) 1.13 A (b) 0.5 A (c) 0.6 A (d) 0.9 A 18. Series AC circuit has inductance L, resistance R and angular frequency w. The quality factor, Q is 2

wL  wL  (a)  (b)  R  R  2 R  R  (c) (d)   wL  wL  19. Current in a circuit is wattless if (a) inductance in the circuit is zero (b) resistance in the circuit is zero (c) current is alternating (d) resistance and inductance both are zero 20. In the circuit shown in figure, the key K is closed at t = 0.

V

K

L

R1 R2

The current through the battery is (a) (b)

V (R1 + R2 ) V at t = 0 and at t = ∞ R1R2 R2 VR1R2 2 1

2 2

at t = 0 and

V at t = ∞ R2

R +R V (R1 + R2 ) V at t = 0 and at t = ∞ (c) R1R2 R2 (d)

VR1R2 V at t = 0 and at t = ∞ R2 R12 + R22 electromagnetic waves

21. The dimensions of (µ0 e0)–1/2 are (a) [L1/2 T–1/2] (b) [L–1 T] –1 (c) [LT ] (d) [L–1/2 T1/2] 22. The charge of a parallel plate capacitor is varying as q = q0 sin2put. The plates are very large and close together. Neglecting edge effects, the displacement current through the capacitor is q q (a) (b) sin2put Ae0 e0 2p u q0 (c) 2puq0 cos2put (d) cos2put. e0 23. Molybdenum is used as a target element for the production of X-rays because it is (a) light and can easily deflect electrons (b) light and can absorb electrons (c) a heavy element with a high melting point (d) an element having high thermal conductivity. 24. The electric field part of an electromagnetic wave in a medium is represented by Ex = 0 ; Ey = 2.5 N C–1 × cos[(2p × 106 rad s–1)t – (p × 10–2 rad m–1)x]; Ez = 0. The wave is (a) moving along the x-direction with frequency 106 Hz and wavelength 100 m (b) moving along x-direction with frequency 106 Hz and wavelength 200 m (c) moving along –x-direction with frequency 106 Hz and wavelength 200 m

(d) moving along y-direction with frequency 2p × 106 Hz and wavelength 200 m. 25. The amplitude of magnetic field in a parallel beam of light of intensity 2.0 W m–2 is (a) 1.3 × 10–7 T (b) 1.3 × 10–6 T –5 (c) 1.3 × 10 T (d) 1.3 × 10–8 T 26. Light with an energy flux of 18 W cm–2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 20 cm2, the total momentum delivered to the surface during a span of 30 minutes is (a) 2.16 × 10–3 kg m s–1 (b) 2.16 × 10–4 kg m s–1 (c) 4.32 × 10–3 kg m s–1 (d) 4.32 × 10–4 kg m s–1 27. If lv, lx and lm represent the wavelengths of visible light, X-rays and microwaves respectively, then (a) lm > lx > lv (b) lv > lm > lx (c) lv > lx > lm (d) lm > lv > lx 28. The electric field associated with an electromagnetic  wave in vacuum is given by E = i 40 cos(kz – 6 × 108 t), where E, z and t are in volt metre–1, metre and second respectively. The value of wave vector k is (a) 2 m–1 (b) 0.5 m–1 –1 (c) 6 m (d) 3 m–1 29. An earth orbiting satellite has solar energy collecting panel with total area 5 m2. If solar radiations are perpendicular and completely absorbed, the average force associated with the radiation pressure is (solar constant = 1.4 kW m–2). (a) 2.33 × 10–3 N (b) 2.33 × 10–4 N –5 (c) 2.33 × 10 N (d) 2.33 × 10–6 N 30. In a region of free space the electric field at some     instant of time is E = (40i + 16 j − 32k) V m–1 and  the magnetic field is B = (0.1i + 0.04 j + 0.145k ) T. The poynting vector for these fields in W m–2 is 7

10 107 [3.6i − 9.0 j] [3.6i − 9.0 j] (b) 2p p 107 107 [3.6i − 9.0 j] [3.6i − 9.0 j] (d) (c) 8p 4p solutions 1. (c) : Current in inductance IL lags behind the emf in phase by p/2, current in resistance IR is in phase with emf, while current in capacitance IC leads by a phase of p/2. (a)

physics for you | MAY ‘16

53

2. (a) : Total resistance of the circuit R = 6 + 4 = 10 W Capacitive reactance 1 1 = XC = = 10 W wC 2000 × 50 × 10−6 Inductive reactance XL = wL = 2000 × 5 × 10–3 = 10 W \ Z = R2 + ( X L − XC )2 = 10 W Amplitude of current V0 20 = I0 = =2A Z 10 p 3. (d) : Current lags behind the voltage by . 2 4. (c) : As, H = I2Rt For DC circuit HDC = I2DCRt …(i) For AC circuit HAC = I2rms Rt …(ii) For same time t, from (i) and (ii) 2 2 I0   H DC  I DC   2 2  = = 2 I rms = =     2 H AC  I rms   2  

5. (a) : As, XL = wL = 2puL = 2p × 750 × 0.1803 = 849.2 W 1 1 1 = 21.2 W and XC = = = wC 2puC 2p × 750 × 10−5 So, X = XL – XC = 849.2 – 21.2 = 828 W Hence, Z = R2 + X 2 = (100)2 + (828)2 = 834 W But as in case of AC, Vrms R × Pav = VrmsIrms cosf = Vrms × Z Z 2 2  Vrms   20  i.e., Pav =   × 100 = 0.0575 W  ×R =  834   Z  And as, U = P × t = mcDq mc × Dq 2 × 10 = = 348 s \ t= P 0.0575 6. (c) : Here, Vp = 2300 V, np = 4000, Vs = 230 V Now, \

Vs ns = Vp n p

ns =

⇒

n 230 = s 2300 4000

230 × 4000 = 400 2300

7. (b) 8. (b) : By introducing the slab, C will increase. Therefore, XC will decrease or I will increase. 9. (a) : The given AC circuit is the combination of two pure parallel circuits with the applied voltage. In which I2 is in phase with V and I1 leads V by 90°. 54

physics for you | MAY ‘16

10. (b) : Here, u = 50 Hz, Irms = 5 A, t =

1 s 300

I0 = 2 Irms = 2 × 5 A From I = I0 sin wt 1 p 3 3 = 5 2 sin 100p × = 5 2 sin = 5 2 × =5 A 300 3 2 2 11. (b) : Pi = 240 × 0.7 = 168 W, P0 = 140 W P 140 η = 0 × 100 = × 100 = 83.3%  80% Pi 168 1 12. (d) : Angular frequency, w = . LC As L1C1 = L2C2 = L3C3, therefore, angular frequency of oscillation is same for all the three circuits. 13. (d) : e = e1 sinwt + e2 coswt The two components are 90° out of phase with each other. e12 + e22

⇒ e0 = ev =

e0 2

=

e12 + e22 2

14. (b) : In RC circuit, the impedance is 1 , as w increases, Z decreases. w2C 2 1 Since, power ∝ , impedance therefore the bulb glows brighter. Z=

R2 +

15. (d) : Here, P = 50 W, V = 100 V P 50 V 100 I= = = 0.5 A, R = = = 200 W V 100 I 0. 5 Let L be the inductance of the choke coil Ev 200 E = = 400 W \ Iv = v or Z = I v 0. 5 Z Now, XL =

Z 2 − R2 = 4002 − 2002 = 100 12

and XL= wL = 100 × 2 3 200 3 200 3 200 3 2 × 1.732 ⇒ L = = 1.1 H = = = w 2pu 100p 3.14 V 200 2 3 = 200 W ; L = H = I 1 p E 200 Z= v = = 400 W I v 0. 5

16. (c) : R =

XL =

Z 2 − R2 = 4002 − 2002 = 200 3

Now, XL = wL = 2puL = 200 3

200 3 200 3 = 50 Hz = 2pL 2 3 2p × p 17. (b) : The current drawn by capacitor and inductor are always in opposite phase. Therefore, net current drawn from the generator = IL – IC = 0.9 – 0.4 = 0.5 A u=

18. (b) : Quality factor of a series LCR circuit is defined as the ratio of the voltage drop across the inductance (or capacitance) at resonance to the applied voltage, wL i.e., Q = R 19. (b) : The current in AC circuit is said to be wattless if the average power consumed in the circuit is zero, i.e., Pav = 0 Thus the current in the circuit has no power. It flows sometimes along the voltage and sometimes against the voltage, so that the net work done per cycle is zero. 20. (c) : At t = 0, inductor (L) offers an infinite resistance, so no current flows through L and R1. V R2 At t = ∞, inductor offers zero resistance, i.e., behaves like a conducting wire. Thus, V (R1 + R2 ) V V = = I= Req R1R2 / (R1 + R2 ) R1R2 Thus, I =

21. (c) :

1

= c (speed of light) and dimensions

µ0 ∈0 of c are [LT–1].

22. (c) : I =

dq d = (q0 sin2put) = q0 2pu cos2put dt dt

23. (c) 24. (b) : Given, Ex = 0, Ey = 2.5 N C–1 × cos[(2p × 106 rad s–1)t – (p × 10–2 rad m–1)x], Ez = 0. This shows that the wave is propagating along x-axis. Comparing the given equation with E = E0 cos (wt – kx), we have w = 2p × 106 or 2pu = 2p × 106 ⇒ u = 106 Hz 2p 2p and = k = p × 10–2 ⇒ l = = 200 m l p× 10−2

25. (a) : Intensity of a wave = or

I = uav c = B0 =

power energy/time = area area

1 2 B0 c ⇒ B0 = 2µ0

2µ0 I c

2 × (4 p × 10−7 ) × 2 = 1.3 × 10–7 T 3 × 108

26. (a) : Total energy falling on the surface = Energy flux × time × area ⇒ U = (18 W cm–2) × (30 × 60 s) × (20 cm2) = 6.48 × 105 J Total momentum delivered to the surface U 6.48 × 105 p= = = 2.16 × 10–3 kg m s–1 c 3 × 108 27. (d) : As wavelength range of microwaves is from 0.1 m to 1mm, visible light is from 700 nm to 400 nm and X-rays is from 1 nm to 10–3 nm \ lm > lv > lx 28. (a) : Given, electric field associated with an electromagnetic wave,  E = i 40 cos(kx – 6 × 108t) Comparing the given equation with E = E0 cos(kx-wt), we have w = 6 × 108 s–1, v = c = 3 × 108 m s–1 w 6 × 108 s −1 k= = = 2 m–1 v 3 × 108 m s −1

29. (c) : As, power = solar constant × area = (1.4 × 103) × 5 solar constant × area \ Force = speed of light 1.4 × 103 × 5 = 2.33 × 10–5 N 3 × 108  1   30. (c) : Poynting vector, P = [E × B] µ0 1    (40i + 16 j − 32k ) × (0.1i + 0.04 j + 0.145k ) = µ0 =

i j k 1 1 [3.6i − 9.0 j] 40 16 −32 = = µ0 4 p× 10−7 0.1 0.04 0.145 =

107 [3.6i − 9.0 j] W m–2 4p physics for you | MAY ‘16

nn 55

1. Which of the following is not dimensionless? (a) Relative density (b) Relative velocity (c) Relative refractive index (d) Relative permittivity

6. The terminal velocity v of a spherical ball of radius r falling through a viscous fluid varies with r such that (a) v/r = constant (b) vr = constant (c) vr2 = constant (d) v/r2 = constant.

2. Two projectiles A and B are projected with angle of projection 30° for the projectile A and 45° for the projectile B. If RA and RB are the horizontal ranges for the two projectiles, then (a) RA = RB (b) RA > RB (c) RA < RB (d) the information is insufficient to decide the relation between RA and RB.

7. At NTP, water boils at 100°C. Deep down a mine, water will boil at a temperature (a) 100°C (b) > 100°C (c) < 100°C (d) will not boil at all.

3. The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M to transfer it from a circular orbit of radius R1 to another of radius R2 (R2 > R1) is  1 1  1 1  (a) GmM  2 − 2  (b) GmM  −   R1 R2   R1 R2  1 1 1 1 1 (c) 2GmM  −  (d) GmM  −  2  R1 R2   R1 R2  4. Wires A and B are made from the same material. A has twice the diameter and three times the length of B. If the elastic limits are not reached when each wire is stretched by the same tension, the ratio of energy stored in A to that in B is (a) 2 : 3 (b) 3 : 4 (c) 3 : 2 (d) 6 : 1 5. At critical temperature, the surface tension of a liquid is (a) zero (b) infinity (c) the same as that at any other temperature (d) cannot be determined. 56

physics for you | MAY ‘16

8. The equation of a progressive wave is given by t x p y = 4 sin p  − +  . Which of the following is 5 9 6  correct? (Assume SI units) (a) v = 5 cm s–1 (b) A = 0.04 cm (c) l = 18 m (d) u = 50 Hz 9. A particle has two equal accelerations in two given directions. If one of the acceleration is halved, then the angle which the resultant makes with the other is also halved. The angle between the accelerations is (a) 120° (b) 90° (c) 60° (d) 45° 10. The amount of work done in stretching a spring from a stretched length of 10 cm to a stretched length of 20 cm is (a) equal to the work done in stretching it from 20 cm to 30 cm (b) less than the work done in stretching it from 20 cm to 30 cm (c) more than the work done in stretching it from 20 cm to 30 cm (d) equal to the work done in stretching it from 0 to 30 cm. 11. A body of mass m strikes a stationary body of mass M and undergoes an elastic collision. After collision,

m has a speed one-third its initial speed. The ratio M/m is (a) 1 : 2 (b) 2 : 1 (c) 1 : 3 (d) 3 : 1 12. A man throws the bricks to a height of 12 m where they reach with a speed of 12 m s–1. If he throws the bricks such that they just reach that height, what percentage of energy will be saved? (g = 9.8 m s–2) (a) 50% (b) 46% (c) 38% (d) 32% 13. A can filled with water is revolved in a vertical circle of radius 4 m and the water just does not fall down. The time period of revolution will be (a) 1 s (b) 10 s (c) 8 s (d) 4 s 14. The moment of inertia of a thin uniform annular disc about one of the diameters is I. Its moment of inertia about an axis perpendicular to the annular surface and passing through its centre is I I (a) 2 I (b) 2I (c) (d) 2 2

(d) both the flux and e.m.f. have their respective maximum values. 19. In Young’s double slit experiment, the central bright fringe can be identified (a) as it has greater intensity than the other bright fringes (b) as it is wider than the other bright fringes (c) as it is narrower than the other bright fringes (d) by using white light instead of monochromatic light. 20. An optical fibre made of glass with a core of refractive index of 1.55 and is clad with another glass with a refractive index of 1.51. Launching takes place from air. What is the value of critical angle for core-clad boundary? (a) 65° (b) 72° (c) 77° (d) 82°

15. A body rolls down an inclined plane. If its kinetic energy of rotational motion is 40% of its kinetic energy of translation, then the body is (a) cylinder (b) ring (c) solid disc (d) solid sphere

21. Heat is flowing through two cylindrical rods of the same material. The diameter of the rods are in the ratio 1 : 2 and the lengths in the ratio 2 : 1. If the temperature difference between the ends be the same, then the ratio of the rates of flow of heat through the two rods will be (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 1 : 8

16. Force acting upon a charged particle kept between the plates of a charge condenser is F. If one of the plates of the condenser is removed then the force acting on the same particle will become (a) zero (b) F/2 (c) F (d) 2F

22. Three moles of oxygen are mixed with two moles of helium. What will be the ratio of specific heats at constant pressure and constant volume for the mixture? (a) 6.76 (b) 1.52 (c) 4.21 (d) 1.48

17. The capacitance of arrangement of 4 plates of area A at a distance d shown in figure, is

23. A tuning fork vibrating with a sonometer having 20 cm wire produces 5 beat s–1. The beat frequency does not change if the length of the wire is changed to 21 cm. The frequency of the tuning fork must be (a) 200 Hz (b) 210 Hz (c) 205 Hz (d) 215 Hz

d A

B

2e 0 A d 4e 0 A (c) 3 e0A/d (d) d 18. When the plane of the armature of an a.c. generator is parallel to the field, in which it is rotating, (a) both the flux linked and induced e.m.f. in the coil are zero (b) the flux linked with it is zero, while induced e.m.f. is maximum (c) flux linked is maximum while induced e.m.f. is zero (a) e0/d

(b)

24. The damping force of an oscillator is directly proportional to the velocity. The unit of the constant of proportionality is (a) kg m s–1 (b) kg m s–2 –1 (c) kg s (d) kg s 25. Ultrasonics are used in SONAR with greatest advantage because ultrasonics (a) have low frequency (b) have short wavelength (c) are electromagnetic waves (d) can be easily produced physics for you | MAY ‘16

57

26. If dielectric constant and dielectric strength be denoted by K and X respectively, then a material suitable for use as a dielectric in a capacitor must have (a) high K and high X (b) high K and low X (c) low K and high X (d) low K and low X 27. The figure shows a parallel circuit, in which the wires have no resistance. As more identical resistors are added to the circuit, the resistance between A and B R R A

B

R

increases and approaches a finite value increases and approaches infinity decreases and approaches zero decreases and approaches a non-zero value.

28. In a region there exist both uniform magnetic field   B and electric field E . An electric charge moving in region experiences no force. This implies that    (a) E, B and v are coplanar    (b) E, B and v are collinear    (c) E = v × B    (d) E = B × v

29. The earth’s magnetic field may be considered to be due to a short magnet placed at the centre of earth and oriented along magnetic south north direction. The ratio of magnitude of magnetic field on earth’s surface at magnetic equator to that at magnetic poles is (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1 30. In the circuit shown in figure, what will be the reading of the voltmeter? V

R

100 V

100 V

L

C

200 V, 100 Hz

58

physics for you | MAY ‘16

(b) 900 V (d) 400 V

31. What is the rms value of an alternating current which when passed through a resistor produces heat, which is thrice that produced by a current of 2 A in the same resistor? (a) 6 A (b) 2 A (c) 3.46 A (d) 0.65 A 32. Electromagnetic radiation of frequency u, of velocity c and wavelength l in air enters a glass slab of refractive index m. The frequency, wavelength and velocity of radiation in glass slab will be c u l (a) , ,c (b) u, l, m m m l c (c) u, , m m

R

(a) (b) (c) (d)

(a) 300 V (c) 200 V

(d)

u l c , , m m m

33. If the ratio of the concentration of electrons to that of holes in a semiconductor is 7/5 and the ratio of current is 7/4, then the ratio of their drift velocities is (a) 5/8 (b) 4/5 (c) 5/4 (d) 4/7 34. A forward biased diode is (a) 0 V –2V (b) – 4 V –3V (c) 3 V 5V (d) – 2 V +2V 35. A signal wave of frequency 12 kHz is modulated with a carrier wave of frequency 2.51 MHz. The upper and lower sideband frequencies are respectively (a) 2512 kHz and 2508 kHz (b) 2522 kHz and 2488 kHz (c) 2502 kHz and 2498 kHz (d) 2522 kHz and 2498 kHz 36. The velocity of the a-particle emitted from uranium of atomic weight 238 is 1.4 × 107 m s–1. The velocity of the remaining nucleus is 1.4 × 107 (a) × 4 m s −1 in the same direction 238 1.4 × 107 (b) × 4 m s −1 in the same direction 234 1.4 × 107 (c) × 4 m s −1 in the opposite direction 234 1.4 × 107 (d) × 4 m s −1 in the opposite direction 238

37. A bubble in glass slab (m = 1.5) when viewed from one side appears at 5 cm and 2 cm from other side, then thickness of slab is (a) 3.75 cm (b) 3 cm (c) 10.5 cm (d) 2.5 cm 38. The wavelength of radiation emitted is l0 when an electron in hydrogen atom jumps from 3rd to 2nd orbit. If in the hydrogen atom itself, the electron jumps from fourth orbit to second orbit, then wavelength of emitted radiation will be 25 27 (a) l (b) l 16 0 20 0 16 20 (c) (d) l l 25 0 27 0 39. Consider two nuclei of the same radioactive nuclide. One of the nuclei was created in a supernova explosion 5 billion years ago. The other was created in a nuclear reactor 5 minutes ago. The probability of decay during the next time is (a) different for each nucleus (b) nucleus created in explosion decays first (c) nucleus created in the reactor decays first (d) independent of the time of creation. 40. The half-life of a radioactive nucleus is 50 days. The time interval (t2 – t1) between the time t2 when (2/3) of it has decayed and the time t1 when (1/3) of it has decayed is (a) 30 days (b) 50 days (c) 60 days (d) 15 days Directions : In the following questions (41-60), a statement of assertion is followed by a statement of reason. Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false.

41. Assertion : For a given time interval, average velocity is single valued while average speed can have many values. Reason : Velocity is a vector quantity and speed is a scalar quantity. 42. Assertion : If a rod having resistance 4 W is turned as half cycle, then its resistance along its diameter is 1.0 W.

Reason : On bending a rod, its length decreases and hence resistance decreases. 43. Assertion : The energy of X-ray photon is greater than that of light photon. Reason : X-ray photon in vacuum travels faster than light photon. 44. Assertion : Refractive index of a medium varies inversely as the temperature of the medium. Reason : Refractive index of a medium varies directly as the density of medium. 45. Assertion : If ice cap of the pole melts, the day length will shorten. Reason : Ice will flow towards the equator and decrease the moment of inertia of the Earth. This increases the frequency of rotation of the Earth. 46. Assertion : Equal masses of helium and oxygen gases are given equal quantities of heat. The rise in temperature of helium is greater than that in the case of oxygen. Reason : The molecular mass of oxygen is more than the molecular mass of helium. 47. Assertion : Two longitudinal waves given by equations : y1 (x, t) = 2a sin (wt – kx), and y2 (x, t) = a sin (2wt – 2kx) will have equal intensity. Reason : Intensity of waves of given frequency in same medium is proportional to the square of the amplitude only. 48. Assertion : When a pendulum is made to oscillate on the surface of the Moon, its time period increases. Reason : Moon is much smaller as compared to Earth. 49. Assertion : On a banked curved track, vertical component of normal reaction provides the necessary centripetal force. Reason : Centripetal force is not always required for turning. 50. Assertion : The logic gate NOT can be built using diode. Reason : The output voltage and input voltage of the diode have 180° phase difference. 51. Assertion : Sky wave signals are used for long distance radio communication. These signals are in general less stable than ground wave signals. physics for you | MAY ‘16

59

Reason : The state of ionosphere varies from hour to hour, day to day and season to season. 52. Assertion : Charge never flows from a condenser of higher capacity to the condenser of lower capacity. Reason : Flow of charge is determined by the charge. 53. Assertion : Absolute zero temperature is also a zero energy temperature of gas molecules. Reason : At absolute zero temperature molecules of a gas come to rest, hence possess no energy of any form.   54. Assertion : If q is the angle between A and B , then   A×B tan q =   . A⋅B     Reason : A × B is perpendicular to A ⋅ B. 55. Assertion : A spaceship while entering the earth’s atmosphere is likely to catch fire. Reason : The temperature of upper atmosphere is very high. 56. Assertion : The sun looks bigger in size at sunrise and sunset than during day. Reason : The phenomenon of diffraction bends light rays. 57. Assertion: Only a change in magnetic flux will maintain an induced current in the coil. Reason: The presence of large magnetic flux through a coil maintains a current in the coil if the circuit is continuous. 58. Assertion : The resistivity of a semiconductor increases with temperature. Reason : The atoms of a semiconductor vibrate with larger amplitude at higher temperatures thereby increasing its resistivity. 59. Assertion : A thin stainless steel needle can lay floating on a still water surface. Reason : Any object floats when the buoyancy force balances the weight of the object. 60. Assertion : Heavy water is a better moderator than normal water. Reason : Heavy water absorbs neutrons more efficiently than normal water. 60

physics for you | MAY ‘16

solutions

1. (b) : Relative velocity has the dimensions of velocity. 2

2. (d) : Horizontal range, R = u sin 2q , g As u for both projectiles A and B are not given, hence relation between RA and RB cannot be decided. 3. (d) : As −

− GMm GMm + KE = 2R1 2R2

1 1  1 KE = GMm  −  . 2  R1 R2  1 4. (b) : Energy stored = × stress × strain × volume 2 1 stress = × stress × × volume 2 Y 1 (stress)2 = × × volume 2 Y \

2

⇒

EA = EB

 F 1  × p(2D/2)2 × 3L  2 2Y  p(2D/2)  2

1  F  × p(D/2)2 × L 2Y  p(D/2)2 

=

3 4

5. (a) 6. (d) 7. (b) : With increase in pressure deep down the mine, boiling point of water increases, i.e., >100°C. t x p 8. (c) :  y = 4 sin p  − +  5 9 6  Comparing this equation with y = A sin (wt – kx + f0), p 2p p k = or = or l = 18 m 9 l 9 B sin q A sin q = (A = B) A + B cos q A + A cos q sin q = ...(i) 1 + cos q β (B / 2)sin q sin q ...(ii) tan = = 2 A + (B / 2)cos q 2 + cos q The equations are satisfied if q = 120° 10. (b) : Work done in stretching a string from 10 cm to 20 cm, 1 1 W = k(x 22 − x12 ) = k(202 − 102 ) = 150k, 2 2 Work done in stretching it from 20 cm to 30 cm 1 W ′ = k(302 − 202 ) = 250 k ⇒ W ′ > W 2 9. (a) : tan β =

u (m − M )u1 (m1 − m2)u1 , \ 1= m+ M 3 m1 + m2 or m + M = 3 m – 3 M M 2 1 or 4 M = 2 m or = = m 4 2 12. (c) : The energy saved is kinetic energy of bricks. KE Percentage of energy saved = ×100% PE + KE 1 2 mv v2 2 = × 100% = × 100% 1 2 gh + v 2 mgh + mv 2 2 122 = × 100% ≈ 38% 2 × 9.8 × 12 + 122 13. (d) : Water does not fall down, when m r w2 = mg g 2p w= = r T 11. (a) : As v1 =

\

T = 2p

r 22 4 =2× =4s g 7 9.8

14. (b) : According to the theorem of perpendicular axes, I′ = IAB + ICD = I + I = 2I C A

B D

KE of rotation 40 2 = = KE of translation 100 5 1 2 1 2 Iw Iw 2 2 2 ⇒ = 2 = or I = mr 2 1 2 1 5 5 mv m r 2w 2 2 2 Hence, the body is a solid sphere. (b) : As one of the plates is removed. E becomes half, therefore, F = qE will also become half i.e. F/2. (b) : The arrangement is equivalent to two capacitors joined in parallel. \ C = 2 e0 A/d (b) : f = NAB cos90° = 0 e = e0 sin90° = e0 = maximum. q is angle between the field and normal to the plane of the coil. (d) : When white light is used instead of monochromatic light, the central bright fringe becomes white, while other are coloured. Hence distinction can be made.

15. (d) :

16. 17.

18.

19.

62

physics for you | MAY ‘16

20. (c) : qc = sin −1  m 2  = sin −1  1.51  = 77°.  m1   1.55  21. (d) : The rate of flow of heat through the rod, H = kA (dT/dx) = k(pD2/4) (T1 – T2)/l, H ∝ D2/l 2

Thus,

2

H1  D1   l2   1   1  1 = = = H 2  D2   l1   2   2  8

22. (d) : g for the mixture is given by n1 + n2 n n = 1 + 2 g − 1 g1 − 1 g 2 − 1 Here, n1 = 3, n2 = 2, g1 = 7/5 and g2 = 5/3 31 = 1.48 21 23. (c) : ... ul = constant (u – 5) 21 = (u + 5)20 or u = 205 Hz hence, we get g =

24. (c) : As F ∝ v, F = kv [kg m s −2] hence k = F = = [kg s −1] v [m s −1] 25. (b) : Since the frequency of ultrasonic waves is greater than audible sound, then their wavelength will be shorter than the audible sound. 26. (a) : The material suitable for use as dielectric must have high dielectric strength X and large dielectric constant K. 27. (c) : The resistors in parallel will provide multiple paths for the current. Due to it, the total resistance will decrease and approach to zero when more resistors in parallel are added. 28. (d) : Total Lorentz force on the charged particle q in the electric and magnetic field is     F = qE + q(v × B)     If F = 0, then qE + q(v × B) = 0      or E = −(v × B) = B × v 29. (a) : Point on magnetic equator is on equatorial line of magnet and at poles, it is on axial line \

Bequatorial 1 / 2 1 = = Baxial 1 2

30. (c) : Here, VL = VC = 100 V, V = 200 V, VR = ? As V = VR2 + (VL − VC )2 \ 200 = VR2 + (100 − 100)2 = VR i.e., VR = 200 V

31. (c) : H1 = 3 H2 ⇒ I2rms R = 3 I2R I rms = I 3 = 2 × 1.732 = 3.46 A 32. (c) : Frequency of radiation does not change with change of medium. Wavelength and velocity of radiation change with change of medium. In medium, wavelength of radiation = l/m velocity of radiation = c/m 33. (c) :

I e neeAv e = = (ne / nh )(v e / v h ) I h nheAv h

v e  I e   nh   7   5  5 = = = v h  I h   ne   4   7  4 (a) : For a p-n junction to be forward-biased, p-side must be at a higher potential than the n-side. (0 V > – 2 V, whereas – 4 V < – 3 V, 3 V < 5 V, – 2 V < + 2V) (d) : USB = 2.51 MHz + 12 kHz = 2510 kHz + 12 kHz = 2522 kHz LSB = 2.51 MHz – 12 kHz = 2510 kHz – 12 kHz = 2498 kHz (c) : a-particle has 4 units mass. Therefore, the mass of remaining nucleus after a-emission = 238 – 4 = 234. If v is the velocity of remaining nucleus, according to law of conservation of linear momentum, we have 238 × 0 = 4 × 1.4 × 107 + 234 × v 4 × 1.4 × 107 or v = − m s −1 234 (c) : Total apparent depth, dA = 5 cm + 2 cm = 7 cm As m = dR/dA, dR = mdA = 1.5 × 7 cm = 10.5 cm or

34.

35.

36.

37.

1  1 1  5R = R 2 − 2  = l0  2 3  36 1  1 1  3R , and = R  2 − 2  = l  2 4  16 l 5R / 36 20 20 \ or l = l 0 = = l 0 3R / 16 27 27 39. (d) : The probability of radioactive decay does not depend on the time of creation of radioactive material. 38. (c) : As

40. (b) : As N = N 0e − lt 2  1  − lt 2 i.e., = e − lt 2  N 0 − 3 N 0  = N 0e 3 1  2  and  N 0 − N 0  = N 0e − lt1 , i.e., = e − lt1 3  3  From equations (i) and (ii) 2 / 3 e − lt1 = or el(t2 – t1) = 2 1 / 3 e − lt 2

ln2 = T1/2 = 50 days l 41. (b) : For a given interval of time, the body in motion may go from one position to another through different paths, where displacement in a given time has one value but distance travelled can have many values. Due to which, assertion is correct. The reason is also correct but not a true explanation of assertion. or l(t2 – t1) = ln 2 or t2 – t1 =

...(i) ...(ii)

42. (d) : When the rod is turned as half cycle there is no change in the dimension of the rod, i.e., length of the rod and its area of cross-section remains the same. Hence, resistance of bent rod is equal to the original resistance. Thus, assertion is wrong. As resistance (R) ∝ l/A, thus reason is also wrong. hc 43. (c) : Energy of a photon E = . As l for X-ray is l smaller than light, so energy of X-rays is greater than that of light. Therefore, assertion is true. Here, reason is false as X-rays and light travel with same speed in vacuum. 44. (b) : With rise in temperature, density of medium usually decreases. Therefore, refractive index of the medium decreases. The reason is true, but it does not explain the assertion correctly. 45. (d) : As the polar ice melts, it spreads out near the axis of rotation of the Earth. On account of this, moment of inertia (I) of Earth increases. Since there is no external torque, L = Iw = I (2p/T) = constant. With increase in I, T will increase, i.e., length of the day will increase. 46. (b) : Heat given to helium is totally used up in increasing the translation energy of its molecules since it is a monoatomic gas. On the other hand, in case of oxygen, which is a diatomic gas, heat is used up in increasing the translational, rotational and vibrational energies of its molecules. The rise in temperature of a gas is due to only increase in translational energy of its molecules. 47. (c) : As intensity, I ∝ w2A2 I1 w 2(2a)2 = =1 I 2 (2w)2(a)2 As apart from A, intensity also depends on w (i.e., frequency u). So, assertion is true, reason is false.

48. (b) : As T = 2p L / g , T increases due to smaller value of g on Moon. physics for you | MAY ‘16

63

49. (d) : Centripetal force is provided by the horizontal component of normal reaction. Vertical component balances the weight of the vehicle. Centripetal force is required for turning. Both, assertion and reason are false. 50. (d) : In diode, the output is in phase with the input. Thus, it cannot be used to build NOT gate. 51. (a) : Since ionospheric properties change with time, sky wave signals are in general less stable than ground wave signals. 52. (d) : Charge flows from a body at higher potential to the body at lower potential. A condenser of higher capacity may be at higher potential. So the charge may flow from it to the condenser of lower capacity, which may be at lower potential. Therefore, assertion and reason are wrong. 53. (d) : At absolute zero temperature, rms velocity of gas molecules becomes zero i.e., molecules cease to translate but they have potential energy. Thus, both assertion and reason are false.    54. (d) : A×B = AB sin θn = tan θn AB cos θ A⋅B   Thus assertion is wrong and (A × B) is a vector quantity and A ⋅ B is a scalar quantity. Thus reason is also wrong. 55. (c) : Here, assertion is true but the reason is false because the spaceship while entering the earth’s atmosphere may catch fire due to atmospheric air friction.

64

physics for you | MAY ‘16

56. (b) : During sunrise and sunset, the ray of light from the sun has to travel a longer distance to reach the eyes. As only longer wavelengths can reach the eyes, the angle subtended by the rays of light on the retina is greater than that during the day. That’s why the sun looks bigger during sunrise and sunset. 57. (c) : Induced current (emf) in a coil is directly proportional to the rate of change of magnetic flux linked with the coil. If there is no change in the flux, there is no induced current also. 58. (d) : With the increase of temperature, the average energy exchanged in a collision increases and so more valence electrons can cross the energy gap, thereby increasing the electron-hole pairs. As in a semiconductor, conduction occurs mainly through electron-hole pairs, so conductivity increases with increase of temperature. Which in turn implies that the resistivity of a semiconductor decreases with rise in temperature. 59. (b) : A thin stainless steel needle on a still water may float due to surface tension as upward force due to surface tension balances the weight of the needle. 60. (c) : The assertion is true because heavy water is a better moderator than ordinary water. This is because heavy water absorbs fewer neutrons than normal water. The reason is false. nn

OLYMPIAD PROBLEMS 1. A submarine made of steel weighing 109 g has to take 108 g of water in order to submerge when the temperature of the sea is 10°C. How much less water it will have to take in when the sea is at 15°C? (Coefficient of cubic expansion of sea water = 2 × 10–4 °C–1, coefficient of linear expansion of steel = 1.2 × 10–5 ºC–1) (a) 1.20 × 102 g (c) 7.91 × 104 g

(b) 3.14 ×103 g

(d) 9.01 × 105 g

2. A ship reaches the entrance to a harbour at noon. The depth of water over the harbour bar is 4 meter at low tide and 10 meter at high tide. Low tide is at 11:20 AM and high tide is at 5:40 PM. Assume that the water surface moves in S.H.M.. The ship requires a depth of 9.4 metre. At what earliest time can the ship cross the bar so as to enter the harbour?

8 times per minute. With what slow vertical speed is the ionospheric layer moving? (Assume the Earth is flat and ignore atmospheric disturbances.) (a) 2.3 cm s–1 (b) 3.2 cm s–1 (c) 2.3 m s–1

(d) 3.2 m s–1

4. Two holes are drilled in the wall of a vessel filled with water. The distances of the holes from the level of water are h and h + H. Find the distances x and y as shown in figure, where the streams flowing out of the holes intersect. Assume that the level of water is maintained in the vessel by regular supply of water. A B C D

x

(a) 12:49 PM (c) 4:18 PM

(b) 2:30 PM (d) 5:16 PM

3. A short wave (HF) radio receiver receives simultaneously two signals from a transmitter 500 km away, one by a path along the surface of the Earth, and one by reflection from a portion of the ionospheric layer situated at a height of 200 km. The layer acts as a perfect horizontal reflector. When the frequency of the transmitted wave is 10 MHz, it is observed that the combined signal strength varies from maximum to minimum and back to maximum

h

x y

H P

y

(a) 2 h (H + h)

H + 2h

(b) 2 H (H + h)

H + 2h

(c) 2 h (H + 2h)

H + 2h

(d) 2 H (H + 2h)

H + 2h

5. A segment of a charged wire of length l, charge density l2, and an infinitely long charged wire, charge density l1, lie in a plane at right angles to each other. The separation between the wires is r0. Determine the force of interaction between the wires. (a) (c)

l1 l2

l   ln 1 +  4 πε0  r0 

l1 l2

l   ln 1 +  2 πε0  r0 

(b) (d)

l1 l2

l   ln 1 −  4 πε0  r0 

l1 l2

l   ln 1 −  2 πε0  r0 

Physics For you | may ‘16

65

++++

solutions

1. (d):

ρwV0g

++++

Shell 6. A solid conducting sphere having a charge Q is Sphere surrounded by an uncharged +++ a Q concentric conducting hollow b spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of hollow shell be V. What will be the new potential difference between the same two surfaces if the shell given a charge –3Q?

+++

(a) V (b) 2V (c) 3V (d) 4V 7. Non-relativistic protons move rectilinearly in the region of space where there are uniform mutually perpendicular electric and magnetic fields. The trajectory of the protons lie in the plane X-Z as shown in figure and forms an angle f with X-axis. Find the pitch of the helical trajectory along which the protons will move after the electric field is switched off.

Mg = ρsV0 g = 109g

Mwg = 108g

rwV0 g = Mw g + Mg rwV0 g = 108 g + 109 g ...(i) rwV0 = 11 × 108 Condition when temperature rised from 10°C to 15°C.

r′w V 0′ g = r′s V 0′ g + mg

(a) (c)

2 π mE qB

2 πmE qB

(b)

2

2

cos f

(d)

2 π mE qB 2 2 πmE qB 2

sin f tan f

8. A metal wire PQ of mass 10 g lies at rest on two horizontal metal rails separated by 5 cm. A vertically downward magnetic field of magnitude 0.800 T exists in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below 20 W, the wire PQ starts sliding on the rails. Find the coefficient of friction.

(a) 0.12 (c) 0.36 66

(b) 0.50 (d) 0.78

Physics For you | may ‘16

...(ii) r′w V0′ g = 109 g + mg rw , V 0′ = V0 (1 + γ s DT ) Now, r′w = (1 + γ w DT ) γ w = 2 × 10−4 °C −1, γ s = 3α s = 3 × 1.2 × 10−5°C −1 DT = 5°C, rwV0 (1 + γ s DT ) \ ...(iii) = 109 + m (1 + γ w DT ) Putting rwV0 from eqn. (i) in (iii), we get 11 × 108 × (1 + γ s DT ) = 109 + m (1 + γ w DT ) ⇒ ⇒

[1 + 5 × 3.6 × 10 −5 ] = 109 + m [1 + 5 × 2 × 10 −4 ] 11 × 108 × (1 + 1.8 × 10 −4 ) = 109 + m (1 + 10 × 10 −4 )

11 × 108 ×

⇒ 11 × 108 + 1.8 × 11 × 104 = 109 + 106 + m + m × 10–3 or 108 – 80.2 × 104 = m (1 + .001) 4 or 10 (10000 – 80.2) = m (1.001) \ m = 99098901.1 So at this position, mass will be 99098901.1 g. So finally reduction in the mass will be = 108 – 99098901.1 = 901098.9 g = 9.01 × 105 g

2. (c) : Time interval between low tide and high tide is half the time period of oscillation. Interval between 11:20 AM and 5:40 PM is 380 minute. Time period = 760 minute. Amplitude of oscillation is 3 m. Centre of oscillation is at a height 7 m above the bar. Water is at half tide point after an interval of T/4 = 190 minute, i.e., at 2:30 PM. x = a sin wt 2π π where, a = 3 m, w = = . 760 380 The ship can cross the bar when the depth is 9.4 m, i.e., x = 2.4 m \ 2.4 = 3 sin wt or sin wt = 0.8 380 −1 ⇒ t = sin (0.8) π = 107.1 minute. Hence the earliest time is 2:30 PM + 1 h 48 min, i.e., at 4:18 PM. 3. (d) : Taking height of ionospheric layer H, length of direct path on the Earth’s surface D, the path difference between two routes is Ionosphere

f =

2Hv u  2 D 2   H +  c  4

v=

fc  2 D 2   H +  2H u  4

1/ 2

(8 / 60) (3 × 108 )  (5 × 105 )2  5 2 2 10 ( × ) +   2 × 2 × 105 × 107  4  –1 = 3.2 m s

1/2

=

4. (a) : At the point P, point of intersection of streams, the range of the two streams are same. As the ranges of liquid issuing from a depth h from the surface and from the base is same, hence CD = AB = h y = H + h + h = H + 2h

or

Time taken by water coming out of C to reach P, 2h g Velocity of water coming out of C, t =

2g (H + h)

Hence required value of x = vt

Transmitter Earth's Receiver surface 1/ 2

−1/2

dH dt

2h × 2 g ( H + h) = 2 h ( H + h) g 5. (c) : Electric field near a long wire is given by l E= 2πε0r The second wire lies in the non-uniform field of first wire. Each element of second wire experiences =

 D2  ...(i) Dx = 2  H 2 +  −D  4  The fluctuation in intensity takes place due to interference between the two signals arriving at the receiver. Each time path difference Dx changes by l (wavelength of the radiation), the received signal strength will vary through one cycle, thus the frequency of the observed fluctuation is 1 d(Dx ) u d (Dx ) ...(ii) f = = l dt c dt where u is the frequency of the radiation and c is speed of light. Differentiating the expression for path difference Dx with respect to time gives

Physics For you | may ‘16

...(iv)

dH is the vertical velocity of the layer. dt On solving the eqn. (iv) for v, we have

v =

D

 d(Dx ) D2  = 2H  H 2 +   4  dt From eqns. (ii) and (iii),

−1/ 2

where v =

H 2 + D2 / 4

H

67

The fluctuation frequency,

...(iii)

Wire 1 λ1

x r0

dx l

Wire 2

“Have the courage to follow your heart and intuition. They somehow know what you truly want to become.” Steve Jobs Physics For you | may ‘16

67

different magnitude of field. Therefore we consider a differential element dx, charge dQ = l2dx, at a distance x from the long wire. The force acting on this element dF is  l  dF = EdQ =  1  l 2 dx  2πε 0 x  The force acting on each element depends on x, the separation between wires 1 and 2. Integrating the expression for dF in the limits x = r0 to x = r0 + l, we obtain r0 + l

F = ∫r

0

l1l 2 dx l1l 2  l  = ln  1 +  2πε0 x 2πε0  r0 

6. (a) : As a and b are the radii of sphere and spherical shell respectively, potential at their surfaces will be, 1 Q Vsphere = 4 πε0 a 1 Q 4 πε0 b And so according to given problem Q 1 1 V = Vsphere − Vshell = − 4 πε0  a b  Now when the shell is given a charge (–3Q) the potential at its surface and also inside will change by Vshell =

and

V0 =

7. (d) : When electric field is switched off, the path followed by the particle will be helical. Pitch = v||T ...(i) where v|| = vcos (90° – f) = vsin f Hence, from eq. (i)  2 πm  ...(ii) Pitch = (v sin f)   qB  When both fields are present qE = qv Bcosf E ...(iii) B cos f Substituting the value of v from eq. (iii) in eq. (ii), we get  2 πm 2 πmE  E = Pitch =  × sin f  tan f  qB  B cos f qB2 8. (a) : The wire starts sliding when magnetic force on the wire overcomes friction force. or

v=

mN = F ⇒ mmg = IlB 6 m × 10 × 10−3 × 10 = × 5 × 10−2 × 0.8 20 or

m=

6 × 5 × 10−2 × 0.8

20 × 10 × 10−3 × 10

= 0.12 

1  −3Q  4 πε0  b 

So, now V′sphere = and V′shell =

1 Q  +V 4 πε0  a  0

1 Q  +V 4 πε0  b  0

and hence V ′sphere − V ′shell =

Q 1 1 − =V 4 πε0  a b 

i.e., if any charge is given to external shell, the potential difference between sphere and shell will not change. This is because by presence of charge on outer shell, potential everywhere inside and on the surface of shell will change by same amount and hence potential difference between sphere and shell will remain unchanged. 68

Physics For you | may ‘16

Physics For you | may ‘16

68

Exam on th 5 June

PRACTICE PAPER 2016 1. A coastguard ship locates a pirate ship at a distance 560 m. It fires a cannon ball with an initial speed 82 m s–1. At what angle from horizontal the ball must be fired so that it hits the pirate ship? (Take g = 10 m s–2) (a) 54° (b) 125° (c) 28° (d) 18° 2. A particle is projected vertically upwards from a point A on the ground. It takes time t1 to reach a point B, but it still continues to move up. If it takes further time t2 to reach the ground from point B. Then height of point B from the ground is 1 (a) g (t1 + t 2 )2 (b) gt1t2 2 1 1 (c) g (t1 + t 2 )2 (d) gt1t 2 2 8 3. A hollow cylinder has a charge q coulomb within it. If f is electric flux in units of volt meter associated with the curved surface B, the flux linked with the plane surface A in units of volt meter will be B A

(a)

q 2e0

C

(b)

f 3

  q  1 q (c)  − f  (d)  − f  2  e0   e0  4. A thin metal disc of radius 0.25 m and mass 2 kg starts from rest and rolls down an inclined plane. If its rotational kinetic energy is 4 J at the foot of the inclined plane, then its linear velocity at the same point is (a) 1.2 m s–1 (b) 2 2 m s −1 (c) 20 m s–1

(d) 2 m s – 1

5. Two radioactive elements X and Y have half lives of 50 minutes and 100 minutes respectively. Initially

both of them have same numbers of atoms. After 200 minutes what is the value of following fraction ? Number of atoms of X unchanged/Number of atoms of Y unchanged. 1 1 (a) 4 (b) 2 (c) (d) 2 4 6. For the circuit shown in figure, the current through the inductor is 0.9 A. While the current through the capacitor is 0.4 A. The current drawn from the generator is C (a) I = 1.13 A L (b) I = 0.9 A (c) I = 0.5 A (d) I = 0.6 A 7. A boy of mass 50 kg is standing on a weighing machine placed on the floor of a lift. The machine reads his weight in N. What is the reading of the machine if the lift is moving upwards with a uniform speed of 10 m s–1? (Take g = 10 m s–2) (a) 510 N (b) 480 N (c) 490 N (d) 500 N 8. Two cells A and B are connected in the secondary circuit of a potentiometer one at a time and the balancing length are respectively 400 cm and 440 cm. The emf of the cell A is 1.08 V. The emf of the second cell B is (a) 1.08 V (b) 1.188 V (c) 11.88 V (d) 12.8 V 9. A simple pendulum is executing SHM with a period of 6 s between two extreme positions B and C about a point O. If the length of the arc BC is 10 cm, how long will the pendulum take the move from position C to a position D towards O exactly midway between C and O. (a) 0.5 s (b) 1 s (c) 1.5 s (d) 3 s 10. The critical angle of a transparent crystal is 45°. Then its polarising angle is Physics For you | may ‘16

69

(a) q = tan −1 ( 2 )

(b) q = sin −1 ( 2 )

 1  −1 (c) q = cos −1   (d) q = cot ( 2 ) 2   11. Two particles of equal mass m go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is Gm Gm (a) (b) R 4R Gm Gm (d) 2R 3R 12. A man standing on a road has to hold his umbrella at 30° with the vertical to keep the rain away. He throws the umbrella and starts running at 10 km h–1. He finds that raindrops are hitting his head vertically. The actual speed of raindrops is (a) 20 km h–1 (b) 10 3 km h −1 (c)

(c) 20 3 km h −1

(d) 10 km h–1

13. The combination of gates shown below yields A X B

(a) NOT gate (c) NAND gate

(b) XOR gate (d) OR gate

14. A mass of 10 kg is suspended from a spring balance. It is pulled a side by a horizontal string so that it makes an angle of 60° with the vertical. The new reading of the balance is (a) 20 kg wt (b) 10 kg wt (c) 10 3 kg wt (d) 20 3 kg wt 15. A garden hose having an internal diameter 2.0 cm is connected to a lawn sprinkler that consists of an enclosure with 24 holes, each 0.125 cm in diameter. If water in the hose has a speed of 90.0 cm s–1, find the speed of the water having the sprinkler hole. (a) 860 cm s–1 (b) 960 cm s–1 –1 (c) 760 cm s (d) 660 cm s–1 16. The radius of a thin wire is 0.16 mm. The area of cross-section of wire in mm2 with correct number of significant figures is (a) 0.081 (b) 0.080 (c) 0.0804 (d) 0.080457 70

Physics For you | may ‘16

17. A transistor is operated in common emitter configuration at VC = 2 V such that a change in the base current from 100 mA to 300 mA produces a change in the collector current from 10 mA to 20 mA. The current gain is (a) 75 (b) 100 (c) 25 (d) 50 18. 1% of 1012 Hz of a satellite link was used from telephony. Find the number of channels (or subscribers) if band width of each channel is 8 kHz. (a) 2.5 × 107 (b) 1.25 × 107 8 (c) 2.5 × 10 (d) 1.25 × 106 19. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 25 revolutions in 14 s, what is the magnitude of acceleration of the stone? (a) 60 m s–2 (b) 100 m s–2 –2 (c) 110 m s (d) 120 m s–2 20. At 27 °C, a gas is suddenly compressed such that 1 its pressure becomes   8

th

of original pressure.

5  Temperature of the gas will be  g =  3  (a) 420 K (b) 327°C (c) 300 K (d) –142°C

21. A solenoid of 0.4 m length with 500 turns carries a current of 3 A. A coil of 10 turns and of radius 0.01 m carries a current of 0.4 A. The torque required to hold the coil with its axis at right angles to that of solenoid in the middle part of it, is (a) 6p2 × 10–7 N m (b) 3p2 × 10–7 N m (c) 9p2 × 10–7 N m (d) 12p2 × 10–7 N m 22. A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4 m s–1. It collides with a horizontal spring of force constant 200 N m–1. The maximum compression produced in the spring will be (a) 0.5 m (b) 0.6 m (c) 0.7 m (d) 0.2 m 23. Two objects of mass 20 g and 30 g are connected by a light rod of length 1 m and move in a horizontal circle as shown in the figure. The speed of each is 2 m s–1. What is the total angular momentum of the objects about the centre? (a) 0.05 kg m2 s–1 (b) 0.01 kg m2 s–1 30 g 20 g (c) 0.15 kg m2 s–1 1m (d) 0.25 kg m2 s–1

24. Electric field on the axis of a small electric dipole at   a distance r is E1 and E2 at a distance of 2r on a line of perpendicular bisector. Then     E E1 (a) E2 = − (b) E2 = − 1 16 8    E1 E (c) E2 = − (d) E2 = 1 4 8 25. The fundamental frequency of a sonometer wire of length l is u0. A bridge is now introduced at a distance Dl(